Energy changes
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Some exam "Dos and Don'ts": Energy changes
Some advice on answering questions on topics often highlighted by examiners as being poorly answered in the past. Make sure that the detail covered by the question is included in your syllabus specification.
Energy in chemical reactions
Exam questions on this topic will usually involve definitions of thermochemical quantities, such as standard enthalpy change of formation, or calculations. Occasionally explanations for differences between calculated quantities, such as bond energies, will be required.
• Know your definitions!
A frequent cause of complaint from examiners - definitions must be precise. Here they must refer to the type of chemical reaction involved, e.g. formation of compound from its elements, complete combustion of a fuel, formation of an ionic lattice from its ions, etc. Specify the amount of substance involved as 1 mole, either formed as product or used as reactant, depending on the definition (hence kJ mol-1 for units). Finally, if 'standard' is included, don't forget to add the standard conditions (298 K, 101 kPa, and 1 M solutions - if applicable) to your definition.
As mentioned elsewhere, it is not a bad idea to give an equation for a reaction that goes with the definition, if there is space. Pay particular attention to state symbols, e.g. Na+(g) + Cl-(g) -> NaCl(s). Leaving them out will lose you the mark.
• Bond energies (enthalpies)
Bond energies merit special mention as their definition often causes problems. Check carefully what your examiners expect. If asked to define average bond energy then you must explain that the average is over bonds of the same type in a range of compounds. The average C-H bond energy will thus differ slightly from the actual bond energy of a particular C-H bond in a given compound.
This is also the reason why enthalpy changes calculated from bond energies do not always agree with the value obtained using enthalpy of formation values. Questions about this difference are usually poorly answered.
Remember too that bond energies involve the making or breaking of bonds where all the species involved are in the gas phase.
• Hess's Law
Constructing a Hess's Law cycle to find the enthalpy change of a reaction (R->P, represented by one side of a triangle) involves finding two reactions which start from the Reactants and Products respectively and reach a common point, thus making up the other two sides of the triangle.
Inevitably the sign of the enthalpy change of one of these reactions will have to be reversed when they are added together to give the required enthalpy change for R->P. A common mistake is to forget to multiply enthalpy changes if more than 1 mole of a substance is involved in the balanced equation.
Typical Hess's Law cycles involve:
1. Decomposition of carbonates: treat both sides with acid to form the same solution in both cases.
2. Hydrated salt -> Anhydrous salt + water: form the same solution from both sides by adding water and use enthplay changes of solution (usually from from experimental data).
3. Combustion: the elements will be common to both sides, so use enthalpies of formation to complete the sides of the triangle.
Another way of dealing with this type problem, especially in multiple-choice questions, is to write down the balanced equation for the reaction you want. Below that write out the balanced equations you have been given data about.
Now reverse one of more of these equations so that when you add them together, they form the equation you want. Treat the enthalpies involved similarly and you'll get the correct answer. For example:
Find enthalpy change for: CS2(g) + 3O2(g) -> CO2(g) + 2SO2(g), given
1. C(s) + 2S(s) -> CS2(g) +110 kJ mol-1
2. C(s) + O2(g) -> CO2(g) -390 kJ mol-1
3. S(s) + O2(g) -> SO2(g) -290 kJ mol-1
(or just the enthalpy of formation values)
CS2 appears as a reactant, so equation 1 must be reversed. Two moles of SO2 are produced, so equation 3 must be doubled.
CS2(g) -> C(s) + 2S(s) -110 kJ mol-1
2S(s) + 2O2(g) -> 2SO2(g) 2(-290) = -580 kJ mol-1
Adding these two equations and equation 2 together should produce the reaction asked for in the question - check this for yourself. Adding the enthlapy changes in the same way gives the required answer: -110 + (-580) + (-390) = -1080 kJ
• Calorimetry calculations
These involve processing data from lab experiments in an insulated cup or vacuum flask (to minimise heat loss) and assume the following:
1. The heat capacity of the mixture in the container is the same as that of water (4.18 J g-1 deg-1)
2. The mass of the mixture heated or cooled by the reaction is equal to the mass of the solutions(s) used, assuming their density equal to that of water (1 g cm-3).
The heat energy change occurring in the container can then be calculated using:
heat energy (J) = mass x specific heat (of water) x temperature change
NOTE: at this stage do
Next scale this 'amount of energy' up to 1 mole of one of the reactants or products, according to the question. The answer should be bigger than the first amount of energy! Convert to kJ.
NOW use your common sense to give your answer a sign, depending on whether the mixture increased in temperature (exothermic) or cooled down (endothermic) and call it 'enthalpy change'.
Many candidates fall at this last fence and get the sign wrong.
• Bond energy calculations
These are usually quite straightforward but many candidates lose marks because they try to apply the formula used for calculating enthalpy changes from enthalpies of formation: Enthalpy change = [Enthalpies of formation of Products] - [Enthalpies of formation of Reactants]
Using bond energies instead of ethalpies of formation inevitably gives an answer with the wrong sign.
Better to tackle such problems this way:
1. First draw all the molecules to show the bonds, if they are not already drawn out this way in the question. This is so you can count all the bonds involved.
2. Bonds broken (in the Reactants): add up all the bonds that are broken when the reactants are converted to separate atoms and add up the energy involved, using the bond energies given. No sign is needed at this stage.
3. Repeat step 2, but for the new bonds formed from separate atoms to make the products. No sign needed here either at this stage.
4. Find the difference between these two amounts of energy - again no sign!
5. Now, remembering that bond breaking requires energy and bond making releases energy, decide if more energy is released making new bonds than is needed to break old bonds here, or vice versa?
If more energy is released, then the reaction is exothermic and now you can give a sign (-) to your answer and call it the 'enthalpy change'. The opposite is of course true if more energy is needed to break the bonds in the reactants than is released when the new bonds are formed. It is safer to use common sense than a formula here!
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updated: 01 June 2007
